If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.

Question

If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.

I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$ But I can't progress after this.

@s1mple From the context follows that it's $xy+4yz+2zx=24.$ See please better the Sujal's attempts. — Michael Rozenberg, Apr 12 '20 at 15:59
@s1mple Sorry, I tend to make typos when I'm typing on a phone. So unfortunately your answer will be wrong. But I take the blame for that. — , Apr 12 '20 at 16:32
@SujalMotagi it's correct, i used what Michael told me. Just check once again. — Nitish Kumar, Apr 12 '20 at 16:34
@s1mple Yeah, I did, it was helpful. I know it's not in your answer, but what does (1,1,1) || (x, 2y, 4z) mean? — , Apr 12 '20 at 16:50

Michael Rozenberg · Answer 1 · 2020-04-12T16:00:07.097

2

Now, by C-S $$144=3\cdot48=(1+1+1)(x^2+4y^2+16z^2)\geq(x+2y+4z)^2=144,$$ where the equality occurs for $$(1,1,1)||(x,2y,4z),$$ which gives $$(x,y,z)=(4,2,1)$$ or $$(x,y,z)=(-4,-2,-1)$$ and $$x^2+y^2+z^2=21.$$

edited Apr 12 '20 at 16:00

answered Apr 12 '20 at 15:54

Michael Rozenberg

203,855

1

You're good at inequalities. This is a brilliant answer. – Nitish Kumar Apr 12 '20 at 16:11
Sorry but what does (1,1,1)||(x,2y,4z) mean? I don't know the notation. – Apr 12 '20 at 16:35
I never expected an inequality in this question. – Apr 12 '20 at 16:37
@Michael Rozenberg, why did you think of using inequalities? I want to improve my problem solving skills so I want to know how you thought of it. – Apr 12 '20 at 16:49
@Sujal Motagi Your problem just looks for using C-S... – Michael Rozenberg Apr 12 '20 at 16:53
@Sujal Motagi $(1,1,1)||(x,2y,4z)$ says that a vector $(1,1,1)$ is parallel to the vector $(x,2y,4z)$, which says that there is $\lambda,$ for which $\lambda(1,1,1)=(x,2y,4z).$ – Michael Rozenberg Apr 12 '20 at 16:56
Ok thanks, I don't know much about vectors. – Apr 12 '20 at 16:58

Nitish Kumar · Accepted Answer · 2020-04-12T16:34:13.893

$x^2+4y^2+16z^2=48 \cdots Eq.(1),\;\; xy+4yz+2zx=24 \cdots Eq.(2)$

From (1), $x^2+(2y)^2+(4z)^2=48 \cdots Eq.(3)$

$(2)\times4 \implies 4xy+16zy+8zx=96\cdots Eq.(4) $

$2\times Eq.(3)-Eq.(4)\implies 2[x^2+(2y)^2+(4z)^2]-2[2xy+8zy+4zx]=0 \implies (x-2y)^2+(2y-4z)^2+(x-4z)^2=0 \implies x=2y=4z.$

Now, $xy+4zy+2zx=24 \implies 3x^2=48\implies x^2=16 $.

Now, $x=4$, then $y=2,z=1$.

$\implies x^2+y^2+z^2=21$.

You will get the same answer even if you take $x=-4$.

score 0 · Answer 3 · answered Apr 12 '20 at 16:50

0

Let $a=x$, $b:=2y$, and $c:=4z$. Then, $$\begin{align}a^2+b^2+c^2&=x^2+4y^2+16z^2=48=2\cdot 24\\&=2\cdot(xy+4yz+2zx)=bc+ca+ab\,.\end{align}$$ It is well known that, for real numbers $a$, $b$, and $c$, $a^2+b^2+c^2=bc+ca+ab$ if and only if $a=b=c$. That is, $$x=2y=4z\,.$$ Since $x^2+4y^2+16z^2=48$, we get $$(x,y,z)=\pm(4,2,1)\,.$$ That is, $$x^2+y^2+z^2=16+4+1=21\,.$$

answered Apr 12 '20 at 16:50

Batominovski

50,341

I didn't know about the property you mentioned. – Apr 12 '20 at 16:59

score 0 · Answer 4 · answered Apr 12 '20 at 20:51

0

$$0 = x^2 + 4 y^2 + 16 z^2 - 2 (4yz + 2zx + xy) = (x-y-2z)^2 + 3 (y-2z)^2 $$ So, as noted $y = 2z$ and $x = y+2z = 4z,$ so $$ 48 = x^2 + 4 y^2 + 16 z^2 = 16 z^2 + 16 z^2 + 16 z^2 = 48 z^2 $$ and $z = \pm 1.$

answered Apr 12 '20 at 20:51

Will Jagy

146,052

If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.

4 Answers4